Optimal. Leaf size=176 \[ \frac {7 (-1)^{3/4} \sqrt {a} \tan ^{-1}\left (\frac {(-1)^{3/4} \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{4 d}+\frac {\tan ^{\frac {3}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}{2 d}-\frac {i \sqrt {\tan (c+d x)} \sqrt {a+i a \tan (c+d x)}}{4 d}+\frac {(1+i) \sqrt {a} \tanh ^{-1}\left (\frac {(1+i) \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{d} \]
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Rubi [A] time = 0.57, antiderivative size = 176, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 9, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.321, Rules used = {3560, 3597, 3601, 3544, 205, 3599, 63, 217, 203} \[ \frac {\tan ^{\frac {3}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}{2 d}+\frac {7 (-1)^{3/4} \sqrt {a} \tan ^{-1}\left (\frac {(-1)^{3/4} \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{4 d}-\frac {i \sqrt {\tan (c+d x)} \sqrt {a+i a \tan (c+d x)}}{4 d}+\frac {(1+i) \sqrt {a} \tanh ^{-1}\left (\frac {(1+i) \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{d} \]
Antiderivative was successfully verified.
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Rule 63
Rule 203
Rule 205
Rule 217
Rule 3544
Rule 3560
Rule 3597
Rule 3599
Rule 3601
Rubi steps
\begin {align*} \int \tan ^{\frac {5}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)} \, dx &=\frac {\tan ^{\frac {3}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}{2 d}-\frac {\int \sqrt {\tan (c+d x)} \left (\frac {3 a}{2}+\frac {1}{2} i a \tan (c+d x)\right ) \sqrt {a+i a \tan (c+d x)} \, dx}{2 a}\\ &=-\frac {i \sqrt {\tan (c+d x)} \sqrt {a+i a \tan (c+d x)}}{4 d}+\frac {\tan ^{\frac {3}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}{2 d}-\frac {\int \frac {\sqrt {a+i a \tan (c+d x)} \left (-\frac {i a^2}{4}+\frac {7}{4} a^2 \tan (c+d x)\right )}{\sqrt {\tan (c+d x)}} \, dx}{2 a^2}\\ &=-\frac {i \sqrt {\tan (c+d x)} \sqrt {a+i a \tan (c+d x)}}{4 d}+\frac {\tan ^{\frac {3}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}{2 d}+i \int \frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {\tan (c+d x)}} \, dx-\frac {(7 i) \int \frac {(a-i a \tan (c+d x)) \sqrt {a+i a \tan (c+d x)}}{\sqrt {\tan (c+d x)}} \, dx}{8 a}\\ &=-\frac {i \sqrt {\tan (c+d x)} \sqrt {a+i a \tan (c+d x)}}{4 d}+\frac {\tan ^{\frac {3}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}{2 d}-\frac {(7 i a) \operatorname {Subst}\left (\int \frac {1}{\sqrt {x} \sqrt {a+i a x}} \, dx,x,\tan (c+d x)\right )}{8 d}+\frac {\left (2 a^2\right ) \operatorname {Subst}\left (\int \frac {1}{-i a-2 a^2 x^2} \, dx,x,\frac {\sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{d}\\ &=\frac {(1+i) \sqrt {a} \tanh ^{-1}\left (\frac {(1+i) \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{d}-\frac {i \sqrt {\tan (c+d x)} \sqrt {a+i a \tan (c+d x)}}{4 d}+\frac {\tan ^{\frac {3}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}{2 d}-\frac {(7 i a) \operatorname {Subst}\left (\int \frac {1}{\sqrt {a+i a x^2}} \, dx,x,\sqrt {\tan (c+d x)}\right )}{4 d}\\ &=\frac {(1+i) \sqrt {a} \tanh ^{-1}\left (\frac {(1+i) \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{d}-\frac {i \sqrt {\tan (c+d x)} \sqrt {a+i a \tan (c+d x)}}{4 d}+\frac {\tan ^{\frac {3}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}{2 d}-\frac {(7 i a) \operatorname {Subst}\left (\int \frac {1}{1-i a x^2} \, dx,x,\frac {\sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{4 d}\\ &=\frac {7 (-1)^{3/4} \sqrt {a} \tan ^{-1}\left (\frac {(-1)^{3/4} \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{4 d}+\frac {(1+i) \sqrt {a} \tanh ^{-1}\left (\frac {(1+i) \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{d}-\frac {i \sqrt {\tan (c+d x)} \sqrt {a+i a \tan (c+d x)}}{4 d}+\frac {\tan ^{\frac {3}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}{2 d}\\ \end {align*}
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Mathematica [A] time = 3.84, size = 267, normalized size = 1.52 \[ \frac {i e^{-i (c+d x)} \left (1+e^{2 i (c+d x)}\right ) \sqrt {a+i a \tan (c+d x)} \left (\frac {\sqrt {2} e^{i (c+d x)} \sqrt {-1+e^{2 i (c+d x)}} \left (1-3 e^{2 i (c+d x)}\right ) \sqrt {\tan (c+d x)}}{\left (1+e^{2 i (c+d x)}\right )^2}+4 \sqrt {2} \sqrt {-\frac {i \left (-1+e^{2 i (c+d x)}\right )}{1+e^{2 i (c+d x)}}} \tanh ^{-1}\left (\frac {e^{i (c+d x)}}{\sqrt {-1+e^{2 i (c+d x)}}}\right )-7 \sqrt {\tan (c+d x)} \tanh ^{-1}\left (\frac {\sqrt {2} e^{i (c+d x)}}{\sqrt {-1+e^{2 i (c+d x)}}}\right )\right )}{4 \sqrt {2} d \sqrt {-1+e^{2 i (c+d x)}}} \]
Antiderivative was successfully verified.
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fricas [B] time = 0.65, size = 559, normalized size = 3.18 \[ \frac {\sqrt {2} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {-i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} {\left (-3 i \, e^{\left (3 i \, d x + 3 i \, c\right )} + i \, e^{\left (i \, d x + i \, c\right )}\right )} + 2 \, {\left (d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )} \sqrt {\frac {49 i \, a}{16 \, d^{2}}} \log \left (\frac {1}{7} \, {\left (7 \, \sqrt {2} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {-i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} {\left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right )} + 8 i \, d \sqrt {\frac {49 i \, a}{16 \, d^{2}}} e^{\left (i \, d x + i \, c\right )}\right )} e^{\left (-i \, d x - i \, c\right )}\right ) - 2 \, {\left (d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )} \sqrt {\frac {49 i \, a}{16 \, d^{2}}} \log \left (\frac {1}{7} \, {\left (7 \, \sqrt {2} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {-i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} {\left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right )} - 8 i \, d \sqrt {\frac {49 i \, a}{16 \, d^{2}}} e^{\left (i \, d x + i \, c\right )}\right )} e^{\left (-i \, d x - i \, c\right )}\right ) - 2 \, {\left (d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )} \sqrt {\frac {2 i \, a}{d^{2}}} \log \left ({\left (\sqrt {2} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {-i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} {\left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right )} + i \, d \sqrt {\frac {2 i \, a}{d^{2}}} e^{\left (i \, d x + i \, c\right )}\right )} e^{\left (-i \, d x - i \, c\right )}\right ) + 2 \, {\left (d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )} \sqrt {\frac {2 i \, a}{d^{2}}} \log \left ({\left (\sqrt {2} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {-i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} {\left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right )} - i \, d \sqrt {\frac {2 i \, a}{d^{2}}} e^{\left (i \, d x + i \, c\right )}\right )} e^{\left (-i \, d x - i \, c\right )}\right )}{4 \, {\left (d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )}} \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [F(-1)] time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [B] time = 0.32, size = 472, normalized size = 2.68 \[ \frac {\sqrt {a \left (1+i \tan \left (d x +c \right )\right )}\, \left (\sqrt {\tan }\left (d x +c \right )\right ) \left (6 i \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {i a}\, \sqrt {-i a}\, \tan \left (d x +c \right )+7 i \ln \left (\frac {2 i a \tan \left (d x +c \right )+2 \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {i a}+a}{2 \sqrt {i a}}\right ) \sqrt {-i a}\, a \tan \left (d x +c \right )+4 i \sqrt {2}\, \ln \left (-\frac {-2 \sqrt {2}\, \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}+i a -3 a \tan \left (d x +c \right )}{\tan \left (d x +c \right )+i}\right ) \sqrt {i a}\, a -4 \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {i a}\, \sqrt {-i a}\, \left (\tan ^{2}\left (d x +c \right )\right )-4 \ln \left (-\frac {-2 \sqrt {2}\, \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}+i a -3 a \tan \left (d x +c \right )}{\tan \left (d x +c \right )+i}\right ) \sqrt {i a}\, \sqrt {2}\, \tan \left (d x +c \right ) a +2 \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {i a}\, \sqrt {-i a}+7 \ln \left (\frac {2 i a \tan \left (d x +c \right )+2 \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {i a}+a}{2 \sqrt {i a}}\right ) \sqrt {-i a}\, a \right )}{8 d \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {i a}\, \sqrt {-i a}\, \left (-\tan \left (d x +c \right )+i\right )} \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \sqrt {i \, a \tan \left (d x + c\right ) + a} \tan \left (d x + c\right )^{\frac {5}{2}}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [F] time = 0.00, size = -1, normalized size = -0.01 \[ \int {\mathrm {tan}\left (c+d\,x\right )}^{5/2}\,\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}} \,d x \]
Verification of antiderivative is not currently implemented for this CAS.
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sympy [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \sqrt {i a \left (\tan {\left (c + d x \right )} - i\right )} \tan ^{\frac {5}{2}}{\left (c + d x \right )}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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